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In the vapor phase, \(\mathrm{BeCl}_{2}\) exists as a discrete molecule. (a) Draw the Lewis structure of this molecule, using only single bonds. Does this Lewis structure satisfy the octet rule? (b) What other resonance structures are possible that satisfy the octet rule? (c) On the basis of the formal charges, which Lewis structure is expected to be dominant for BeCl_ \(_{2} ?\)

Short Answer

Expert verified
The Lewis structure for BeCl2 with single bonds is: Cl - Be - Cl. However, the octet rule is not satisfied for Be, as it only has 4 electrons. No resonance structures can satisfy the octet rule. The formal charges for Be and Cl atoms are 0 in the current structure, which makes this single-bonded structure the dominant one for BeCl2.

Step by step solution

01

Draw the Lewis Structure of BeCl2 with Single Bonds

First, let's draw the Lewis structure of BeCl2 using only single bonds: 1. Calculate the total number of valence electrons: Be has 2 valence electrons and each Cl has 7, so in total there are \(2 + 2(7) = 16\) valence electrons. 2. Place the least electronegative element, Be, in the center, and Cl atoms on both sides. 3. Connect each Cl atom to Be with a single bond (2 electrons per bond, 4 total). 4. Distribute the remaining 12 electrons to Cl atoms, filling their octets. Now, we have the following structure: ``` Cl - Be - Cl ```
02

Check if the Octet Rule is Satisfied

With the current Lewis structure, both Cl atoms have a complete octet with 8 electrons. However, the Be atom has only 4 electrons (2 single bonds with Cl atoms). Therefore, the octet rule is not satisfied for Be.
03

Identify Possible Resonance Structures

To satisfy the octet rule for Be, we can think of possible resonance structures. We can try moving one electron pair from a Cl atom to Be, creating a double bond between Be and Cl, which would give Be an octet. However, this would also result in Be exceeding its valence limit (maximum of 2 electrons), which is not favorable. Thus, no resonance structures can satisfy the octet rule.
04

Calculate Formal Charges

We will now calculate the formal charges to determine the most dominant Lewis structure: 1. For Be: Formal charge of Be = \(2 - (0 + 4/2) = 0\) 2. For Cl: Formal charge of Cl = \(7 - (6 + 1/2) = 0\) In our current Lewis structure, both Be and Cl atoms have a formal charge of 0.
05

Determine Dominant Lewis Structure

Since there are no resonance structures that satisfy the octet rule and the formal charges for the current structure are 0 for all atoms, we can conclude that this single-bonded Lewis structure is the dominant one for BeCl2: ``` Cl - Be - Cl ```

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Octet Rule
The octet rule is a guiding principle in chemistry that suggests atoms seek to have eight electrons in their outermost shell to achieve stability, resembling the electron configuration of a noble gas. However, in practice, some elements do not strictly adhere to this rule, particularly those in earlier periods like beryllium (Be).

For \( \text{BeCl}_{2} \), the Lewis structure involves placing beryllium at the center attached to two chlorine atoms, each with a single bond. Cl atoms satisfy the octet rule with eight electrons around them, including the bonding ones. In contrast, beryllium is stable with only four electrons due to its small size and position on the periodic table, which deviates from the octet expectation.

While the octet rule provides a straightforward way to predict the distribution of electrons in molecules, exceptions occur frequently, particularly for elements beyond the second period or with atomic numbers less than eight.
Formal Charge
Formal charge is a concept used to determine the most stable Lewis structure among possible resonance forms. It is calculated using the formula:\[ \text{Formal Charge} = \text{Valence Electrons} - (\text{Lone Pair Electrons} + \frac{1}{2}\text{Bonding Electrons}) \]

In the case of \( \text{BeCl}_{2} \), the formal charges were calculated as 0 for both beryllium and chlorine atoms. Beryllium has two valence electrons, and with four bonding electrons and no lone pairs, its formal charge is:\[ 2 - (0 + \frac{4}{2}) = 0 \]

Similarly, chlorine has seven valence electrons, and with six electrons in lone pairs and one from bonding, its formal charge is also:\[ 7 - (6 + \frac{2}{2}) = 0 \]

Structures with formal charges closest to zero tend to be the most stable, explaining why the simple form with single bonds remains the favored structure.
  • Use formal charge calculations to validate structures.
  • Structures with minimized formal charges (preferably zero) are typically the most stable.
Resonance Structures
Resonance structures are different ways of arranging electrons in a molecule that can't be represented by a single Lewis structure. They help depict the delocalization of electrons within molecules.

For \( \text{BeCl}_{2} \), it initially seems like creating double bonds might form alternative resonance structures. However, attempting this would lead Be to exceed its typical electron capacity, moving beyond its available valence shell capability.

Thus, no valid resonance structures exist for \( \text{BeCl}_{2} \) under standard oxidation state rules. Resonance structures are generally applicable to systems with delocalized electrons, as seen in molecules like benzene or ions like sulfate.

In summary:
  • Resonance structures are alternate forms illustrating electron distribution.
  • Not all molecules have feasible resonance structures.
  • Ensure electron count and formal charges are logical in any attempted resonance form.

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Most popular questions from this chapter

(a) Draw the Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\) . (b) What is the weakest bond in hydrogen peroxide? (c) Hydrogen peroxide is sold commercially as an aqueous solution in brown bottles to protect it from light. Calculate the longest wavelength of light that has sufficient energy to break the weakest bond in hydrogen peroxide.

Based on Lewis structures, predict the ordering, from shortest to longest, of \(N-O\) bond lengths in \(N O^{+}, N O_{2}^{-},\) and \(N O_{3}^{-} .\)

True or false: (a) The \(\mathrm{C}-\) Cbonds in benzene are all the same length and correspond to typical single \(\mathrm{C}-\mathrm{Cbond}\) lengths. (b) The \(\mathrm{C}-\mathrm{C}\) bond in acetylene, HCCH, is longer than the average \(\mathrm{C}-\mathrm{C}\) bond length in benzene.

Although \(\mathrm{I}_{3}\) is a known ion, \(\mathrm{F}_{3}^{-}\) is not. (a) Draw the Lewis structure for \(\mathrm{I}_{3}^{-}\) (it is linear, not a triangle). (b) One of your classmates says that \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{Fis}\) too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong. (c) Another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because it would violate the octet rule. Is this classmate possibly correct? (d) Yet another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{F}\) is too small to make bonds to more than one atom. Is this classmate possibly correct?

Ammonia reacts with boron trifluoride to form a stable compound, as we saw in Section 8.7 . (a) Draw the Lewis structure of the ammonia-boron trifluoride reaction product. (b) The B-N bond is obviously more polar than the \(\mathrm{C}-\mathrm{C}\) bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus and delta minus symbols mentioned in Section 8.4\()\) . ( ) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(B-N\) bond in the trichloride reaction product would be more or less polar than the \(B-N\) bond in the trifluoride product, and justify your reasoning.

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